Python concurrent.futures.wait job submission order not preserved

Does python's concurrent.futures.wait() preserve the order of job submission? I submitted two jobs to ThreadPoolExecutor as follows:

import concurrent.futures
import time, random

def mult(n):
    time.sleep(random.randrange(1,5))
    return f"mult: {n * 2}"

def divide(n):
    time.sleep(random.randrange(1,5))
    return f"divide: {n // 2}"


with concurrent.futures.ThreadPoolExecutor() as executor:
    mult_future = executor.submit(mult, 200)
    divide_future = executor.submit(divide, 200)

    # wait for both task to complete
    mult_task, divide_task = concurrent.futures.wait(
        [mult_future, divide_future],
        return_when=concurrent.futures.ALL_COMPLETED,
    ).done

    mult_result = mult_task.result()
    divide_result = divide_task.result()

    print(mult_result)
    print(divide_result)

Sometimes I see

divide: 50
mult: 400

and sometimes,

mult: 400
divide: 50

shouldn't mult_task, divide_task always map to mult_future, divide_future ?

python --version
>> Python 3.8.16

According to the doc that the function wait():

Returns a named 2-tuple of sets.

By definition, sets are unordered. That means you cannot guarantee the order.

Since you already have mult_future and divide_future, you can use them to guarantee the order. There is no need to call wait either.

import concurrent.futures
import random
import time


def mult(n):
    time.sleep(random.randrange(1, 5))
    return f"mult: {n * 2}"


def divide(n):
    time.sleep(random.randrange(1, 5))
    return f"divide: {n // 2}"


with concurrent.futures.ThreadPoolExecutor() as executor:
    mult_future = executor.submit(mult, 200)
    divide_future = executor.submit(divide, 200)

# Print in order: mult div
print(mult_future.result())
print(divide_future.result())